Question

若x^2-3x+1=0,求代数式4x^2-[3x^2+2(x^2-x)-4x-5]的值

快点，明天要交了

Answer 1

x^2-3x+1=0, (x-1)(2x-1)=0, x1=1, x2=1/2

4x^2-[3x^2+2(x^2-x)-4x-5]
=4x^2-[3x^2+2x^2-2x-4x-5]
=4x^2-[5x^2-6x-5]
=4x^2-5x^2+6x+5
=-x^2+6x+5
=-(x^2-6x-5)
①=-(x^2-6x-5)
=-[1-6-5]
=10
②=-(x^2-6x-5)
=-[1/4-3-5]
=8-1/4
=7又3/4

Answer 2

15加减6根号2 我计算不太好，作为参考吧